Question

According to a report in USAToday, more and more parents are helping their young adult children get homes. Suppose eight persons in a random sample of 40 young adults who recently purchased a home in Kentucky received help from their parents. You have been asked to construct a 95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents. What is the margin of error for a 95% confidence interval for the population proportion?

Answer 1

Answer:

a)  95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents.

( 0.0761 , 0.3239)

b) Margin of error  =  0.1264.

Step-by-step explanation:

Explanation:-

Given '8' persons in a random sample of 40 young adults who recently purchased a home in Kentucky received help from their parents.

sample proportion of success $'p' = \frac{8}{40} = 0.2$

q = 1=p

q = 1-0.2 = 0.8

a)

95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents.

$(p-1.96\sqrt{\frac{pq}{n} } , p+1.96\sqrt{\frac{pq}{n} } )$

$(0.2-1.96\sqrt{\frac{0.2X0.8}{40} } , 0.2+1.96\sqrt{\frac{0.2X0.8}{40} } )$

(0.2 - 0.1239,0.2+0.1239)

( 0.0761 , 0.3239)

b) the margin of error for a 95% confidence interval for the population proportion.

For the 95% confidence interval ∝= 0.05 and zₐ = 1.96≅2.

$Margin of error = \frac{2\sqrt{pq} }{\sqrt{n} }$

$Margin of error = \frac{2\sqrt{0.2X0.8} }{\sqrt{40} }$

Margin of error for a 95% confidence interval for the population proportion.

Margin of error  =  0.1264.